Compute battery life from mAh and current draw?

Question

I have a circuit that operates at about 7amps. I need it to operate at that current for about 15 minutes. How do I compute how long a battery will last based on its mAH rating?

For example, say I have a battery rated at 1800 mAh. How long with that battery last in a circuit drawing 7amps?

Thanks

Answer 1

An amp is 1000 milliamps, so your circuit draws 7000 milliamps. You need power for 15 minutes, which is 0.25 hours. So you simply multiply.

7000 ma * 0.25 hrs = 1750 milliamp-hrs.

Technically, your 1800 ma-hr battery could handle it. But, as with most manufacturers' specifications, the rating can be, shall we say, a bit optimistic. I would want to see more margin for error, at least 2000 ma-hr, if not 2500, especially for critical systems. On the other hand, if power failure near the end is not critical, then the 1800 ma-hr battery should get you most of the way there.

Answer 2

The mAh rating is just the product of current and time. Your circuit will run for 1/4 hour at 7000 mA for 1750 mAh. It would be nice if that were all. So far, I've only covered the battery energy. You need to deal with the power as well. Just like different types of battery provide different voltages and energy, they have different power ratings as well. A current draw of 7 amps will greatly reduce tie voltage from a typical alkaline C or D cell and even from a large lantern battery. To get that kind of current from an alkaline, you need put several batteries in parallel.

A more suitable solution is to use a high power chemistry. These are batteries designed for rapid discharge. Common high-power batteries are nickel cadmium (Ni-Cad), nickel metal hydride (NiMH), lead-acid and some of the new high power lithiums. Even a C-cell of Ni-Cad or NiMH will easily provide 7 amps for 15 minutes. You could probably even do it with 2 AA cells in parallel. Keep in mind that you need enough batteries in series to give the needed voltage as well as enough in parallel (maybe just one) to give enough current.

Answer 3

7 amps is 7000 ma so 1800 mAH battery should last just 15 minutes IF it is fully charged and relatively new. I would use a 20% safety factor on that and get a bigger battery. Also if the current draw has peaks I would derate the battery further.

Another factor to consider is the battery type as the ability to maintain output voltage as it discharges varies greatly with type. And your circuit's performance at lower voltages is also a factor.

Answer 4

each and every of the solutions are terrific, yet you should verify the size of the burden on the battery and time you like it to function. If looking strictly at how plenty skill you are able to draw, whether merely for a short volume of time... a Thevenin equivilant circuit is your terrific guess to degree the output obtainable. In Thevein you tournament your load on your source, then degree your load to be sure the source's skill. connect the battery to a variable load (resistance) like a extensive potentiometer or decade field of resistors. placed your volt meter around the burden and advance the burden until the voltage reads one one million/2 of the battery voltage. Disconnect the circuit. degree the resistance of the burden, and persist with ohms regulation. because one million/2 the voltage is interior the burden/one million/2 interior the source you have an identical impedence, and as such, an identical skill. Volts divided by making use of Ohms = contemporary. that provides you with the theoretical terrific output you are able to ever get from that battery. How long it is going to final relies upon on the size of the battery (and can merely be got here across making use of a provide up watch).